Schrödinger-Operator on $L^2[0,2\pi]$.

Question

In Reed-Simon Analysis of Operators they often talk about operators like $H = - \Delta +V$ as an operator on $L^2[0,2\pi]$ (like in Theorem XIII 88. What do they mean by that? Or is their a canonical dense subset that is meant instead?-Of course, this operator is not well-defined on $L^2[0,2\pi]$, so what is their intention behind this?

Answer 1

You're thinking of a domain $\mathscr{D}(H)$ which is dense in $X=L^{2}$. So they really mean $$ H : \mathscr{D}(H) \subset X \rightarrow X. $$ The domain $\mathscr{D}(H)$ is where the appropriate endpoint conditions are imposed. So the domain might be, for example, the space of twice absolutely-continuous functions $f$ on $[0,2\pi]$ for which $-f''+Vf \in L^{2}$ and such that $f(0)=f(2\pi)$. The more you restrict $H$, the larger the domain of the adjoint $H^{\star}$ becomes. You want to find just the right balance so that $H=H^{\star}$, meaning that both domains are identical (which is the meaning of selfadjoint for unbounded operators.) The Mathematical Foundations of Quantum Mechanics were first defined this way by John von Neumann in the 1930's, who also formulated and proved the Spectral Theorem for unbounded densely-defined selfadjoint linear operators on a complex Hilbert space.

Answer 2

$L^2$ is nice because you have the inner product. What they probably mean is some Hilbert Space ($H^s =L^2$ such that $s$ derivatives make sense, note that this is just a very informal definition). So we have that $S = - \Delta + V$ acts as follows:

$$S:H^s \to L^2$$

Notice that if we're working with $H^s$ everything will make sense if $s\geq 2$ since we know that $\psi \in H^2 \implies \Delta \psi \in L^2$